Question

two persons A & B are located in XY plane at the points (0,0) & (0,10) resp. (the distance are measured in MKS units). at a time t=0, they start moving simultaneously with velocities Va=2j m/s
& Vb=2i m/s ,resp, the time aftr which A & B are at their closest distance is.....
1. 2.5 s
2. 4 s
3. 1 s
4. 10/1.414

Answer 1

@Michele_Laino

Answer 2

@IrishBoy123

Answer 3

hi

i'd start with position vectors \(\vec r_A(t) = <0, 2t>, \quad \vec r_B(t) = <2t, 10>\) so \(\vec r_{BA} (t) = <0-2t, 2t-10> = \dots\)

so \( | \vec r_{BA} |^2= 4t^2 + ( 2t-10)^2 = \dots\)

complete the algebra and then solve for \(\dfrac{d (| \vec r_{BA} |^2)}{dt} = 0\)

you can also compute \(\dfrac{d^{\color{red}{2}} ( | \vec r_{BA} |^2)}{dt ^ {\color{red}{2}} }\) to establish it as a min but that is kinda obvious from the physical set up.....

Answer 4

hey....tysm

Answer 5

bt y u differntiate it...???

Answer 6

by diff it we get velocityt...!!!

Answer 7

nah. here you are differentiating the square of the distance between the particles.


to elaborate: \[d^2 = | \vec r_{BA} |^2= 4t^2 + ( 2t-10)^2 = \dots\]

and so you look at \[\dfrac{d (d^2)}{dt} = 0\]

Answer 8

bt why???
sry bt i couldnt undrstnd this!!!

Answer 9

sorry pratima, my bad

we know that at time t the coordinates of A and B are given by A = (0, 2t) and B = (2t, 10). Do you see that?

using Pythagoreas, we know that the distance between them at time t is given by \(s(t) = \sqrt{(0-2t)^2 + (2t-10)^2} = \sqrt{8t^2 - 40t + 100}\)

to find the minumum distance, we need to calculate \(\dfrac{ds}{dt} = \dfrac{\frac{1}{2} (16t - 40)}{\sqrt{8t^2 - 40t + 100}} = 0\)

\(\implies 16t - 40 = 0, \implies t = { 5 \over 2}\) sec.

i was suggesting calculating the derivative of the square of the distance because the algebra is a bit easier and it gives the same answer. but i think i just confused.

hope that's clearer :-)

Answer 10

ty ty....its clear nw....ty vry much :) :)